Answer
Verified427.8k+ views
Hint: In this question, we are given roots of the equation ${{x}^{2}}+px+q=0$ and we need to find roots of the equation $\left( {{p}^{2}}-4q \right)\left( {{p}^{2}}{{x}^{2}}+4px \right)-16q=0$. For this, we will find values of p and q using sum of roots and product of roots formula and then we will simplify the equation whose roots are to be found. After simplifying we will be able to find roots of the equation easily. For an equation $a{{x}^{2}}+bx+c$ sum of roots is given by $\dfrac{-b}{a}$ and product of roots is given by $\dfrac{c}{a}$.
Complete step-by-step solution
Here, we are given the roots of the equation ${{x}^{2}}+px+q=0$ as $\left( \alpha +\sqrt{\beta } \right)\text{ and }\left( \alpha -\sqrt{\beta } \right)$. We need to find roots of the equation $\left( {{p}^{2}}-4q \right)\left( {{p}^{2}}{{x}^{2}}+4px \right)-16q=0$. For this, let us find values of p and q.
Now, we know that, for an equation $a{{x}^{2}}+bx+c$ sum of roots is equal to $\dfrac{-b}{a}$ and product of roots is equal to $\dfrac{c}{a}$.
Comparing ${{x}^{2}}+px+q=0$ we get the sum of roots as -p and product of roots as q.
Since, $\left( \alpha +\sqrt{\beta } \right)\text{ and }\left( \alpha -\sqrt{\beta } \right)$ are roots of the equation ${{x}^{2}}+px+q$. So,
$\left( \alpha +\sqrt{\beta } \right)+\left( \alpha -\sqrt{\beta } \right)=-p\Rightarrow 2\alpha =-p\Rightarrow p=-2\alpha $.
Also, $\left( \alpha +\sqrt{\beta } \right)\left( \alpha -\sqrt{\beta } \right)=q$.
Applying $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ on above equation, we get:
${{\alpha }^{2}}-{{\left( \sqrt{\beta } \right)}^{2}}=q\Rightarrow {{\alpha }^{2}}-\beta =q$.
Hence values of p and q are $-2\alpha \text{ and }\left( {{\alpha }^{2}}-\beta \right)$ respectively.
We need to find roots of the equation $\left( {{p}^{2}}-4q \right)\left( {{p}^{2}}{{x}^{2}}+4px \right)-16q=0$. So let us put values of p and q to simplify equation, we get:
\[\Rightarrow \left( {{\left( -2\alpha \right)}^{2}}-4\left( {{\alpha }^{2}}-\beta \right) \right)\left( {{\left( -2\alpha \right)}^{2}}{{x}^{2}}+4\left( -2\alpha \right)x \right)-16\left( {{\alpha }^{2}}-\beta \right)=0\]
Simplifying and opening brackets we get:
\[\begin{align}
& \Rightarrow \left( 4{{\alpha }^{2}}-4{{\alpha }^{2}}+4\beta \right)\left( 4{{\alpha }^{2}}{{x}^{2}}-8\alpha x \right)-16{{\alpha }^{2}}+16\beta =0 \\
& \Rightarrow \left( 4\beta \right)\left( 4{{\alpha }^{2}}{{x}^{2}}-8\alpha x \right)-16{{\alpha }^{2}}+16\beta =0 \\
& \Rightarrow 16{{\alpha }^{2}}\beta {{x}^{2}}-32\alpha \beta x-16{{\alpha }^{2}}+16\beta =0 \\
\end{align}\]
Taking 16 common from all terms and taking it to the other side, we get:
\[\Rightarrow {{\alpha }^{2}}\beta {{x}^{2}}-2\alpha \beta x-{{\alpha }^{2}}+\beta =0\]
Now taking $\alpha \beta $ common from first two terms, we get:
\[\begin{align}
& \Rightarrow \alpha \beta \left( \alpha {{x}^{2}}-2x \right)-{{\alpha }^{2}}+\beta =0 \\
& \Rightarrow \alpha \beta \left( \alpha {{x}^{2}}-2x \right)={{\alpha }^{2}}-\beta \\
\end{align}\]
Dividing both sides by $\alpha \beta $, we get:
\[\Rightarrow \left( \alpha {{x}^{2}}-2x \right)=\dfrac{{{\alpha }^{2}}-\beta }{\alpha \beta }\]
Now for finding value of x, let us complete the square of left side of the equation, for this, let us first take $\alpha $ common, we get:
\[\Rightarrow \alpha \left( {{x}^{2}}-\dfrac{2}{\alpha }x \right)=\dfrac{{{\alpha }^{2}}-\beta }{\alpha \beta }\]
Now we know that, for completing square in ${{x}^{2}}+bx+c$ we need to add and subtract ${{\left( \dfrac{b}{2} \right)}^{2}}$ term.
So, for above equation, we will add and subtract ${{\left( \dfrac{2}{2\alpha } \right)}^{2}}={{\left( \dfrac{1}{\alpha } \right)}^{2}}$.
Hence equation becomes \[\Rightarrow \alpha \left( {{x}^{2}}-\left( \dfrac{2}{\alpha } \right)x+{{\left( \dfrac{1}{\alpha } \right)}^{2}}-{{\left( \dfrac{1}{\alpha } \right)}^{2}} \right)=\dfrac{{{\alpha }^{2}}-\beta }{\alpha \beta }\].
Now, ${{x}^{2}}-\left( \dfrac{2}{\alpha } \right)x+{{\left( \dfrac{1}{\alpha } \right)}^{2}}$ is of the form ${{a}^{2}}+{{b}^{2}}-2ab$ hence, its equal to ${{\left( a-b \right)}^{2}}$ we get:
\[\Rightarrow \alpha \left( {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}-{{\left( \dfrac{1}{\alpha } \right)}^{2}} \right)=\dfrac{{{\alpha }^{2}}-\beta }{\alpha \beta }\]
Taking $\alpha $ to other side, we get:
\[\begin{align}
& \Rightarrow {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}-{{\left( \dfrac{1}{\alpha } \right)}^{2}}=\dfrac{{{\alpha }^{2}}-\beta }{{{\alpha }^{2}}\beta } \\
& \Rightarrow {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}=\dfrac{{{\alpha }^{2}}-\beta }{{{\alpha }^{2}}\beta }+\dfrac{1}{{{\alpha }^{2}}} \\
\end{align}\]
Now let us simplify right side by taking ${{\alpha }^{2}}\beta $ as LCM we get:
\[\begin{align}
& \Rightarrow {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}=\dfrac{{{\alpha }^{2}}-\beta +\beta }{{{\alpha }^{2}}\beta } \\
& \Rightarrow {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}=\dfrac{{{\alpha }^{2}}}{{{\alpha }^{2}}\beta } \\
& \Rightarrow {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}=\dfrac{1}{\beta } \\
\end{align}\]
Taking square root both sides we get:
\[\Rightarrow x-\dfrac{1}{\alpha }=\pm \dfrac{1}{\sqrt{\beta }}\]
Hence
\[\Rightarrow x-\dfrac{1}{\alpha }=\dfrac{1}{\sqrt{\beta }}\text{ and }x-\dfrac{1}{\alpha }=-\dfrac{1}{\sqrt{\beta }}\].
Solving these two equation we get value of x as,
\[\begin{align}
& \Rightarrow x=\dfrac{1}{\sqrt{\beta }}+\dfrac{1}{\alpha }\text{ and }x=-\dfrac{1}{\sqrt{\beta }}+\dfrac{1}{\alpha } \\
& \Rightarrow x=\dfrac{1}{\alpha }+\dfrac{1}{\sqrt{\beta }}\text{ and }x=\dfrac{1}{\alpha }-\dfrac{1}{\sqrt{\beta }} \\
\end{align}\]
Hence, option A is the correct answer.
Note: Students should note that calculations in this question are very complex so take care while performing them. Students can make the mistake of forgetting negative terms when taking the square root. Take care of signs while using the formula of the sum of roots and product of roots. Make sure to first divide and then square the middle term while completing the square.
Complete step-by-step solution
Here, we are given the roots of the equation ${{x}^{2}}+px+q=0$ as $\left( \alpha +\sqrt{\beta } \right)\text{ and }\left( \alpha -\sqrt{\beta } \right)$. We need to find roots of the equation $\left( {{p}^{2}}-4q \right)\left( {{p}^{2}}{{x}^{2}}+4px \right)-16q=0$. For this, let us find values of p and q.
Now, we know that, for an equation $a{{x}^{2}}+bx+c$ sum of roots is equal to $\dfrac{-b}{a}$ and product of roots is equal to $\dfrac{c}{a}$.
Comparing ${{x}^{2}}+px+q=0$ we get the sum of roots as -p and product of roots as q.
Since, $\left( \alpha +\sqrt{\beta } \right)\text{ and }\left( \alpha -\sqrt{\beta } \right)$ are roots of the equation ${{x}^{2}}+px+q$. So,
$\left( \alpha +\sqrt{\beta } \right)+\left( \alpha -\sqrt{\beta } \right)=-p\Rightarrow 2\alpha =-p\Rightarrow p=-2\alpha $.
Also, $\left( \alpha +\sqrt{\beta } \right)\left( \alpha -\sqrt{\beta } \right)=q$.
Applying $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ on above equation, we get:
${{\alpha }^{2}}-{{\left( \sqrt{\beta } \right)}^{2}}=q\Rightarrow {{\alpha }^{2}}-\beta =q$.
Hence values of p and q are $-2\alpha \text{ and }\left( {{\alpha }^{2}}-\beta \right)$ respectively.
We need to find roots of the equation $\left( {{p}^{2}}-4q \right)\left( {{p}^{2}}{{x}^{2}}+4px \right)-16q=0$. So let us put values of p and q to simplify equation, we get:
\[\Rightarrow \left( {{\left( -2\alpha \right)}^{2}}-4\left( {{\alpha }^{2}}-\beta \right) \right)\left( {{\left( -2\alpha \right)}^{2}}{{x}^{2}}+4\left( -2\alpha \right)x \right)-16\left( {{\alpha }^{2}}-\beta \right)=0\]
Simplifying and opening brackets we get:
\[\begin{align}
& \Rightarrow \left( 4{{\alpha }^{2}}-4{{\alpha }^{2}}+4\beta \right)\left( 4{{\alpha }^{2}}{{x}^{2}}-8\alpha x \right)-16{{\alpha }^{2}}+16\beta =0 \\
& \Rightarrow \left( 4\beta \right)\left( 4{{\alpha }^{2}}{{x}^{2}}-8\alpha x \right)-16{{\alpha }^{2}}+16\beta =0 \\
& \Rightarrow 16{{\alpha }^{2}}\beta {{x}^{2}}-32\alpha \beta x-16{{\alpha }^{2}}+16\beta =0 \\
\end{align}\]
Taking 16 common from all terms and taking it to the other side, we get:
\[\Rightarrow {{\alpha }^{2}}\beta {{x}^{2}}-2\alpha \beta x-{{\alpha }^{2}}+\beta =0\]
Now taking $\alpha \beta $ common from first two terms, we get:
\[\begin{align}
& \Rightarrow \alpha \beta \left( \alpha {{x}^{2}}-2x \right)-{{\alpha }^{2}}+\beta =0 \\
& \Rightarrow \alpha \beta \left( \alpha {{x}^{2}}-2x \right)={{\alpha }^{2}}-\beta \\
\end{align}\]
Dividing both sides by $\alpha \beta $, we get:
\[\Rightarrow \left( \alpha {{x}^{2}}-2x \right)=\dfrac{{{\alpha }^{2}}-\beta }{\alpha \beta }\]
Now for finding value of x, let us complete the square of left side of the equation, for this, let us first take $\alpha $ common, we get:
\[\Rightarrow \alpha \left( {{x}^{2}}-\dfrac{2}{\alpha }x \right)=\dfrac{{{\alpha }^{2}}-\beta }{\alpha \beta }\]
Now we know that, for completing square in ${{x}^{2}}+bx+c$ we need to add and subtract ${{\left( \dfrac{b}{2} \right)}^{2}}$ term.
So, for above equation, we will add and subtract ${{\left( \dfrac{2}{2\alpha } \right)}^{2}}={{\left( \dfrac{1}{\alpha } \right)}^{2}}$.
Hence equation becomes \[\Rightarrow \alpha \left( {{x}^{2}}-\left( \dfrac{2}{\alpha } \right)x+{{\left( \dfrac{1}{\alpha } \right)}^{2}}-{{\left( \dfrac{1}{\alpha } \right)}^{2}} \right)=\dfrac{{{\alpha }^{2}}-\beta }{\alpha \beta }\].
Now, ${{x}^{2}}-\left( \dfrac{2}{\alpha } \right)x+{{\left( \dfrac{1}{\alpha } \right)}^{2}}$ is of the form ${{a}^{2}}+{{b}^{2}}-2ab$ hence, its equal to ${{\left( a-b \right)}^{2}}$ we get:
\[\Rightarrow \alpha \left( {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}-{{\left( \dfrac{1}{\alpha } \right)}^{2}} \right)=\dfrac{{{\alpha }^{2}}-\beta }{\alpha \beta }\]
Taking $\alpha $ to other side, we get:
\[\begin{align}
& \Rightarrow {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}-{{\left( \dfrac{1}{\alpha } \right)}^{2}}=\dfrac{{{\alpha }^{2}}-\beta }{{{\alpha }^{2}}\beta } \\
& \Rightarrow {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}=\dfrac{{{\alpha }^{2}}-\beta }{{{\alpha }^{2}}\beta }+\dfrac{1}{{{\alpha }^{2}}} \\
\end{align}\]
Now let us simplify right side by taking ${{\alpha }^{2}}\beta $ as LCM we get:
\[\begin{align}
& \Rightarrow {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}=\dfrac{{{\alpha }^{2}}-\beta +\beta }{{{\alpha }^{2}}\beta } \\
& \Rightarrow {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}=\dfrac{{{\alpha }^{2}}}{{{\alpha }^{2}}\beta } \\
& \Rightarrow {{\left( x-\dfrac{1}{\alpha } \right)}^{2}}=\dfrac{1}{\beta } \\
\end{align}\]
Taking square root both sides we get:
\[\Rightarrow x-\dfrac{1}{\alpha }=\pm \dfrac{1}{\sqrt{\beta }}\]
Hence
\[\Rightarrow x-\dfrac{1}{\alpha }=\dfrac{1}{\sqrt{\beta }}\text{ and }x-\dfrac{1}{\alpha }=-\dfrac{1}{\sqrt{\beta }}\].
Solving these two equation we get value of x as,
\[\begin{align}
& \Rightarrow x=\dfrac{1}{\sqrt{\beta }}+\dfrac{1}{\alpha }\text{ and }x=-\dfrac{1}{\sqrt{\beta }}+\dfrac{1}{\alpha } \\
& \Rightarrow x=\dfrac{1}{\alpha }+\dfrac{1}{\sqrt{\beta }}\text{ and }x=\dfrac{1}{\alpha }-\dfrac{1}{\sqrt{\beta }} \\
\end{align}\]
Hence, option A is the correct answer.
Note: Students should note that calculations in this question are very complex so take care while performing them. Students can make the mistake of forgetting negative terms when taking the square root. Take care of signs while using the formula of the sum of roots and product of roots. Make sure to first divide and then square the middle term while completing the square.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE